Physics • Year 12 • Module 5 • Lesson 2
Launch Angle Analysis
Apply the range equation and complementary angle property to data tables, graphs, and scenario-based reasoning questions.
1. Interpret a range data table — shot put investigation
A physics class investigated how launch angle affects range. They used a launcher that always fires at 12.0 m/s (g = 9.8 m/s², level ground). Their results are shown below. 8 marks
| Launch angle (θ) | sin(2θ) | Measured range (m) | Calculated range R = v² sin(2θ)/g (m) | % difference |
|---|---|---|---|---|
| 15° | 0.500 | 7.2 | ||
| 30° | 0.866 | 12.5 | ||
| 45° | 1.000 | 14.5 | ||
| 60° | 0.866 | 12.4 | ||
| 75° | 0.500 | 7.1 |
1.1 Complete the “Calculated range” column for each angle. Show one full calculation below. 3 marks
1.2 Calculate the percentage difference between measured and calculated range for θ = 45°. Use: % difference = |measured − calculated| / calculated × 100. What does a non-zero percentage difference suggest about real projectile motion? 2 marks
1.3 Identify which pairs of angles give the same calculated range and explain why using the complementary angle property. 3 marks
2. Interpret a range vs launch angle graph
The graph below shows range as a function of launch angle for a projectile fired at 20.0 m/s on level ground (g = 9.8 m/s²). 7 marks
Figure 2.1. Range vs launch angle for v = 20.0 m/s on level ground, g = 9.8 m/s². Theoretical values only (no air resistance).
2.1 Describe the shape of the curve and identify the launch angle that gives the maximum range. Read off the maximum range from the graph. 2 marks
2.2 The graph appears symmetrical about 45°. Choose one pair of complementary angles (e.g. 30° and 60°) and verify from the graph that they give the same range. Then explain mathematically why this must always be true. 3 marks
2.3 A student claims: “At 90° launch angle, the range is zero because there is no horizontal velocity.” Use the range equation to verify this claim and confirm whether the claim is correct. 2 marks
3. Compare launch from level ground vs launch from a cliff
Complete the table below, contrasting key features of projectile launch on level ground with launch from a 40 m cliff. 10 marks (1 per cell)
| Feature | Level ground launch | Cliff launch (h = 40 m above landing) |
|---|---|---|
| Vertical displacement at landing (sy) | ||
| Time-of-flight formula | ||
| Optimal angle for max range | ||
| Maximum range formula (exact) | ||
| Do complementary angles give equal ranges? |
4. Predict and justify — a javelin throw scenario
An athlete throws a javelin from shoulder height (release point is 1.8 m above the ground). The athlete can throw at 28 m/s regardless of angle. 5 marks
4.1 Predict whether the optimal launch angle for maximum range will be greater than, less than, or equal to 45°. Justify your prediction using the concept of launch height and flight time. 3 marks
4.2 A sports scientist claims that world-record javelin throws use launch angles between 33° and 36°, not 45°. Is this consistent with your prediction? Identify one additional real-world factor (beyond launch height) that might also shift the optimal angle below 45°. 2 marks
Q1.1 — Calculated ranges (v² = 144; g = 9.8)
R = 144 × sin(2θ) / 9.8. Values: 15°: 144 × 0.500 / 9.8 = 7.35 m; 30°: 144 × 0.866 / 9.8 = 12.7 m; 45°: 144 × 1.000 / 9.8 = 14.7 m; 60°: 12.7 m; 75°: 7.35 m.
Q1.2 — Percentage difference at 45° and its implication
% difference = |14.5 − 14.7| / 14.7 × 100 ≈ 1.4%. A non-zero difference suggests that real conditions (air resistance, finite launch height, rotation of the projectile) cause measured ranges to fall below the theoretical vacuum prediction. Award 1 mark for correct calculation, 1 mark for a valid physical explanation.
Q1.3 — Complementary pairs
15° and 75° are complementary (15 + 75 = 90°); both give sin(30°) = 0.500, hence R ≈ 7.35 m. Similarly, 30° and 60° are complementary (30 + 60 = 90°); both give sin(60°) = 0.866, hence R ≈ 12.7 m. This follows from the identity sin(180° − 2θ) = sin(2θ): replacing θ with (90° − θ) gives sin(2(90° − θ)) = sin(180° − 2θ) = sin(2θ). Award 1 mark per pair identified, 1 mark for the mathematical explanation.
Q2.1 — Shape and maximum
The curve is a smooth, symmetric arch. It rises from 0 at 0°, reaches a single peak at 45° (R = 40.8 m), then falls symmetrically back to 0 at 90°. The maximum range is 40.8 m (read from the peak). Award 1 mark for describing the symmetrical arch shape peaking at 45°, 1 mark for reading 40.8 m.
Q2.2 — Complementary pair verification and explanation
Example: at 30°, graph reads ≈ 35.4 m; at 60°, graph reads ≈ 35.4 m — equal. Mathematically: R(30°) = 400 sin(60°)/9.8 = 400 × 0.866/9.8 = 35.4 m; R(60°) = 400 sin(120°)/9.8 = 400 × 0.866/9.8 = 35.4 m. Since sin(120°) = sin(180° − 60°) = sin(60°), complementary pairs always give the same range. Award 1 mark for graph read-off, 1 mark for correct calculation, 1 mark for explicit mathematical identity argument.
Q2.3 — Verification at 90°
R = v² sin(2 × 90°) / g = v² sin(180°) / g = v² × 0 / g = 0. The range is zero. The student’s claim is correct: at 90° the projectile goes straight up, so vx = v cos(90°) = 0 and no horizontal distance is covered. Award 1 mark for correct substitution and result, 1 mark for physical interpretation.
Q3 — Compare and contrast table
sy at landing: Level = 0 m (same height). Cliff = −40 m (40 m below launch). Time of flight: Level = 2v sinθ / g. Cliff = (v sinθ + √(v² sin²θ + 2gh)) / g. Optimal angle: Level = 45°. Cliff = less than 45° (typically 33–40° depending on h). Rmax formula: Level = v²/g (exact, at θ = 45°). Cliff = no closed-form simplification; found numerically. Complementary angles give equal ranges? Level = Yes. Cliff = No; the height asymmetry breaks the sin(2θ) symmetry.
Q4.1 — Optimal angle for javelin
The optimal angle is less than 45°. Because the javelin is released 1.8 m above the ground, it lands at a lower height than it was released. This extra downward displacement extends the flight time compared to a level-ground launch. A shallower angle preserves a larger horizontal velocity component (vx = v cosθ), which acts over the extended flight time to produce greater horizontal range. Award 1 mark for correct prediction, 2 marks for the reasoning linking launch height to extended flight time and larger horizontal velocity component.
Q4.2 — Consistency with real data and extra factor
Yes, 33°–36° is consistent: the optimal angle is below 45° because the release point is above the ground. An additional real-world factor is air resistance — drag reduces range, and aerodynamic forces on the javelin (lift from its orientation) shift the optimum further. Also valid: the athlete cannot achieve equal launch speed at all angles (biomechanical speed-angle trade-off reduces force output at steep angles). Award 1 mark for confirming consistency, 1 mark for any valid additional factor with brief explanation.